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Exercise 5.4.20

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If $U = g (T)$ , we know that $UT = TU$ since

T (T^{k}) = (T^{k}) T = T^{k + 1}

and $T$ is linear. For the converse, we may suppose that $V$ is generated by $v$ . Then the set

β = {v, T (v), T^{2} (v), \dots, T^{k} (v)}

is a basis. So the vector $U (v)$ could be written as a linear combination of $β$ . This means $U (v) = g (T) (v)$ for some polynomial $g$ . Now if $UT = TU$ , we want to show $U = g (T)$ by showing $U (w) = g (T) (w)$ for all $w \in β$ . Observe that

U (T^{m} (v)) = T^{m} (U (v)) = T^{m} g (T) (v) = g (T) (T^{m} (v))

for all nonnegative integer $m$ . So we get the desired result.

jephianlin

2011-06-27 00:00

Exercise 5.4.20

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